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Proof step 3 from proof int2bits-truein-01 (previous step)

2 ^ i + j = 0, true ∈ int2bits(j).rest, 0 ≤ i ⊦ j = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: j = ~(2 ^ i)

Additional information

Used simplifier rules: int2bits-2pot-01: 0 ≤ i ⊦ int2bits(~(2 ^ i)) = true ' + mklist(false, i→n(i));% ⊦ ~ i = j ↔ i + j = 0;% ⊦ ~ i + i = 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ~ i = 0 ↔ i = 0;% ⊦ 0 ≤ j → 2 ^ j ≠ 0;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ true ∈ mklist(false, n); Used formulas: j = ~(2 ^ i)